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arsen [322]
3 years ago
14

Are all decimals a rational number

Mathematics
2 answers:
ivanzaharov [21]3 years ago
5 0
No not all since for example 5.0 is in decimal form but is not a rational number
kow [346]3 years ago
3 0
Only if it's not a whole number by itself, like 3.0 or 8.0 but yeah pretty much
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198 is 33% of what number?<br> Please show work<br> Can Robtobey help me and others :) ...?
Ghella [55]
First, you have to convert 33% to decimals by dividing it to 100. And that would be 0.33.
Then you want to know the 100 percent of 198 because it's just a 33% part of a whole.
You will have to divide 198 by 33% or 0.33 to get the whole. So the answer would be 600.
8 0
3 years ago
Please help me to find the %
Viktor [21]
1.068 * 60 - 18.39 =?
8 0
3 years ago
Ms. Gallegos burns 236 calories riding her bike each hour. She wants to burn more than 590 calories riding her bike at the same
mrs_skeptik [129]

Answer:

Bro the answer is A

Step-by-step explanation:

trust me straight 100% brain dead person

6 0
3 years ago
IQ scores are normally normally distributed with a mean of 100 and center deviation of 15 if one person is randomly selected wha
Arlecino [84]

Answer:

22.86% probability that the persons IQ is between 110 and 130

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 100, \sigma = 15

If one person is randomly selected what is the probability that the persons IQ is between 110 and 130

This is the pvalue of Z when X = 130 subtracted by the pvalue of Z when X = 110.

X = 130

Z = \frac{X - \mu}{\sigma}

Z = \frac{130 - 100}{15}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 110

Z = \frac{X - \mu}{\sigma}

Z = \frac{110 - 100}{15}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

0.9772 - 0.7486 = 0.2286

22.86% probability that the persons IQ is between 110 and 130

6 0
3 years ago
How do you answer this?
lidiya [134]
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3 
(a +(- b))^3 = (a-b)^2 =  a^3 - 3a^2b + 3ab^2- b^3  
7 0
3 years ago
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